已知cos2a=4/5,则cos^4 a+sin^4 a=
\u5df2\u77e5cos2a=3/5\u5219sin^4+cos^4\u7684\u503c\u4e3a\u591a\u5c11\uff1fcos2a=3/5
2(cosa)^2-1=1-2(sina)^2=3/5
(cosa)^2=4/5,(sina)^2=1/5
sin^4+cos^4=[(sina)^2+(cosa)^2]^2-2(sina)^2*(cosa)*2
=1-2*1/5*4/5
=17/25
\u6211\u4eec\u77e5\u9053\uff0csin\u5e73\u65b9a+cos\u5e73\u65b9a=1\uff0c\u5c06\u4e24\u8fb9\u5e73\u65b9\u5f97\uff0ccos4\u6b21a+sin4\u6b21a+2sin\u5e73\u65b9a*cos\u5e73\u65b9a=1\uff0c\u6240\u4ee5cos4\u6b21a+sin4\u6b21a=1-2sin\u5e73\u65b9a*cos\u5e73\u65b9a...........\u2474\uff0c
\u800c2sin\u5e73\u65b9acos\u5e73\u65b9a=\uff08sin\u5e73\u65b92a\uff09/2....................\u2475\uff0c
sin\u5e73\u65b92a=1-\uff08cos\u5e73\u65b92a\uff09=1-\uff0816/25\uff09=9/25\uff0c\u4ee3\u5165\u2475\u4e2d\uff0c\u5f972sin\u5e73\u65b9a*cos\u5e73\u65b9a=9/50..........................\u2476\uff0c\u5c06\u2476\u4ee3\u5165\u2474\u4e2d\uff0c\u5f97\uff0ccos4\u6b21a+sin4\u6b21a=1-\uff089/50\uff09=41/50
将两边平方得，cos4次a+sin4次a+2sin平方a*cos平方a=1，
所以cos4次a+sin4次a=1-2sin平方a*cos平方a...........⑴，
而2sin平方acos平方a=（sin平方2a）/2....................⑵，
sin平方2a=1-（cos平方2a）=1-（16/25）=9/25，
代入⑵中，得2sin平方a*cos平方a=9/50..........................⑶，
将⑶代入⑴中，得，cos4次a+sin4次a=1-（9/50）=41/50
一道三角函数习题已知cos2a=4/5,则cos4次a+sin4次a=??? 需要具体的我们知道，sin平方a+cos平方a=1，将两边平方得，cos4次a+sin4次a+2sin,
绛旓細鍏紡鍟婏紝tana=sina/cosa=2锛屾湁锛坰ina锛塣2/(cosa)^2=4,(sina)^2=0.8,(cosa)^2=0.2,sin2a=2sinacosa=2骞虫柟鏍癸紙0.8*0.2锛=2*0.4=0.8 cos2a=骞虫柟鏍癸紙1-0.8^2锛=0.6 tan2a=0.8/0.6=4/3
绛旓細宸茬煡COS2A=鍥鍒嗕箣涓锛屽垯SIN2A=鍏垎涔嬩笁
绛旓細瑙ｉ鎬濊矾锛氱洿鎺ュ埄鐢ㄤ簩鍊嶈鐨勪綑寮﹀嚱鏁板叕寮忥紝姹傚嚭sin 2伪鐨勫硷紝寰楀嚭閫夐」锛巆os2伪=1-2sin2伪锛屸埓[1/4]=1-2sin2伪锛屸埓sin2伪=[3/8]锛屾晠閫塂锛,2,锛2009•骞垮窞妯℃嫙锛宸茬煡cos2伪=[1/4]锛屽垯sin 2伪=锛堬級A锛嶽1/2]B锛嶽3/4]C锛嶽5/8]D锛嶽3/8]
绛旓細璇烽噰绾
绛旓細=4[sin(90-A/2)]^2-cos2A =4(cosA/2)^2-cos2A =4[(1+cosA)/2]-cos2A =2+2cosA-2(cosA)^2+1 =-2(cosA)^2+2cosA+3=7/2 4(cosA)^2-4cosA+1=0 (2cosA-1)^2=0 2cosA=1 cosA=1/2 A=60掳锛孊+C=120掳锛宑os锛圔+C锛=cosBcosC-sinBsinC=-1/2,sinBsinC=1/2+...
绛旓細cos4娆℃柟a+sin4娆℃柟a+sin²acos²a =cos4娆℃柟a+2sin²acos²a+sin4娆℃柟a-sin²acos²a =(sin²a+cos²a)²-sin²acos²a =1-1/4sin²2a =1-1/4(1-cos²2a)=49/64 ...
绛旓細2伪鈭(0,蟺/2)鎵浠in2a>0 cos2a>0 tan2a=4/3 sin2a/cos2a=4/3 浠in2a=4t cos2a=3t 鍥犱负sin²2a+cos²2a=1 16t²+9t²=1 t²=1/25 t=1/5鎴-1/5锛堣垗鍘伙級鏍规嵁棰樻剰 sin2a=4/5 cos2a=3/5 ...
绛旓細cos2a=1-2sina骞虫柟 sina骞虫柟=锛1-cos2a锛/2 =3/8
绛旓細瑙 cos2a=1-2sin²a=1/4 鈭 2sin²a=3/4 鈭磗in²a=3/8
绛旓細cos2伪=cos²伪-sin²伪 =cos²伪-sin²伪/1 =cos²伪-sin²伪/cos²伪+sin²伪 涓婁笅鍚屾椂闄や互cos²伪 寰1-tan²伪/1+tan²伪 绛変簬-15/17
